If I_{n} = int_{0}^{frac{pi}{4}} tan^{n} x , | vedclass

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(D) $I_{n} = \int_{0}^{\frac{\pi}{4}} 	an^{n} x \, dx$ ... $(i)$ $I_{n+2} = \int_{0}^{\frac{\pi}{4}} 	an^{n+2} x \, dx$ ... $(ii)$ Adding equations

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$\frac{1}{9}$

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(D) $I_{n} = \int_{0}^{\frac{\pi}{4}} 	an^{n} x \, dx$ ... $(i)$ $I_{n+2} = \int_{0}^{\frac{\pi}{4}} 	an^{n+2} x \, dx$ ... $(ii)$ Adding equations $(i)$ and $(ii)$: $I_{n} + I_{n+2} = \int_{0}^{\frac{\pi}{4}} 	an^{n} x (1 + 	an^{2} x) \, dx$ $I_{n} + I_{n+2} = \int_{0}^{\frac{\pi}{4}} 	an^{n} x \sec^{2} x \, dx$ Let $u = 	an x$,then $du = \sec^{2} x \, dx$. When $x = 0, u = 0$ and when $x = \frac{\pi}{4}, u = 1$. $I_{n} + I_{n+2} = \int_{0}^{1} u^{n} \, du = \left[ \frac{u^{n+1}}{n+1} ight]_{0}^{1} = \frac{1}{n+1}$. For $n = 8$,we have $I_{8} + I_{10} = \frac{1}{8+1} = \frac{1}{9}$.

If $I_{n} = \int_{0}^{\frac{\pi}{4}} \tan^{n} x \, dx$,where $n$ is a positive integer,then $I_{10} + I_{8}$ is equal to

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